展开两个向量列

Question

我有一个有两列的tibble。每列包含有序的纬度、经度对。结构如下：

library(dplyr)
library(tidyr)

> my_df
# A tibble: 3 x 2
  V1        V2       
  <list>    <list>   
1 <dbl [2]> <dbl [2]>
2 <dbl [2]> <dbl [2]>
3 <dbl [2]> <dbl [2]>

my_df = structure(list(V1 = list(c(44.0252714, -88.1536451), c(42.9856117, 
-87.9355419), c(42.8600366, -87.9541568)), V2 = list(c(44.9535298, 
-90.9188588), c(45.4864422, -89.7339536), c(43.0743635, -87.9765372
))), row.names = c(NA, -3L), class = c("tbl_df", "tbl", "data.frame"
))

我想把它变成一个四列的数据框，看起来像：

> my_df2
        y1        x1       y2        x2
1 44.02527 -88.15365 44.95353 -90.91886
2 42.98561 -87.93554 45.48644 -89.73395
3 42.86004 -87.95416 43.07436 -87.97654

我曾尝试使用tidyr 中的unnest，但没有成功。

> my_df %>% unnest()
# A tibble: 6 x 2
     V1    V2
  <dbl> <dbl>
1  44.0  45.0
2 -88.2 -90.9
3  43.0  45.5
4 -87.9 -89.7
5  42.9  43.1
6 -88.0 -88.0

> my_df %>% unnest(V1, V2)
# A tibble: 6 x 2
     V1    V2
  <dbl> <dbl>
1  44.0  45.0
2 -88.2 -90.9
3  43.0  45.5
4 -87.9 -89.7
5  42.9  43.1
6 -88.0 -88.0

我需要以某种方式控制取消嵌套的发生方式，但我不知道如何去做。

Answer 1

这是一个技巧，您首先将每个向量转换为字符串：

my_df %>%
  rowwise() %>%
  mutate_all(funs(toString(.))) %>%
  separate(V1, c("y1", "x1"), ", ") %>%
  separate(V2, c("y2", "x2"), ", ") %>%
  mutate_all(funs(as.numeric(.)))
# A tibble: 3 x 4
     y1    x1    y2    x2
  <dbl> <dbl> <dbl> <dbl>
1  44.0 -88.2  45.0 -90.9
2  43.0 -87.9  45.5 -89.7
3  42.9 -88.0  43.1 -88.0

编辑 更类似于 base-R 的方法：

my_df2 <- 
  do.call(cbind, lapply(my_df, function(x) do.call(rbind, x))) %>% 
  as.tibble()
names(my_df2) <- c("y1", "x1", "y2", "x2")
my_df2
# A tibble: 3 x 4
     y1    x1    y2    x2
  <dbl> <dbl> <dbl> <dbl>
1  44.0 -88.2  45.0 -90.9
2  43.0 -87.9  45.5 -89.7
3  42.9 -88.0  43.1 -88.0

Answer 2

一个可能的解决方案是这样的：

library(dplyr)
library(tidyr)

my_df %>% 
  rowwise() %>%                                  # for each row
  mutate_all(funs(list(data.frame(t(.))))) %>%   # transpose your vector and create a dataframe
  unnest()                                       # unnest

# # A tibble: 3 x 4
#      X1    X2   X11   X21
#   <dbl> <dbl> <dbl> <dbl>
# 1  44.0 -88.2  45.0 -90.9
# 2  43.0 -87.9  45.5 -89.7
# 3  42.9 -88.0  43.1 -88.0

Answer 3

与data.table::transpose...

rn_df = function(x, suff = 0, cols = c("x","y"))
  setNames(x, paste0(cols, suff))

my_df %>% 
  lapply(data.table::transpose) %>%
  unname %>%
  Map(rn_df, ., seq_along(.)) %>%
  unlist(recursive=FALSE) %>% 
  data.frame

        x1        y1       x2        y2
1 44.02527 -88.15365 44.95353 -90.91886
2 42.98561 -87.93554 45.48644 -89.73395
3 42.86004 -87.95416 43.07436 -87.97654

这应该扩展到原始my_df 中的一些列，假设它们都变成具有相同命名模式的列。

---

在我看来，最好使用长格式的数据，而不是使用稍后必须解析的 col 名称：

library(data.table)
res = my_df %>% lapply(. %>% (data.table::transpose) %>% setDT) %>%
  rbindlist(id = "src") %>%
  setnames(-1, c("x", "y"))

   src        x         y
1:  V1 44.02527 -88.15365
2:  V1 42.98561 -87.93554
3:  V1 42.86004 -87.95416
4:  V2 44.95353 -90.91886
5:  V2 45.48644 -89.73395
6:  V2 43.07436 -87.97654